Question

Let \( A_1, A_2, A_3, \dots, A_9 \) be 9 AM’s between 49 and 149. Then the mean of \( A_1, A_2, A_3 \) and \( A_9 \) is:

• A. 110
• B. 120
• C. 130
• D. 140

Hint:

The sum of all $n$ AMs is $n \times (\frac{a+b}{2})$. This is a useful shortcut for questions asking for the sum of all inserted means.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
We have 49 Arithmetic Means between \( a = 49 \) and \( b = 149 \). Including the endpoints, there are \( n + 2 = 51 \) terms in the Arithmetic Progression.
Step 2: Key Formula or Approach:
The common difference \( d \) is given by:
\[ d = \frac{b - a}{n + 1} \]
where \( n = 49 \).
Step 3: Detailed Explanation:
\[ d = \frac{149 - 49}{49 + 1} = \frac{100}{50} = 2 \]
The \( k \)-th Arithmetic Mean is \( A_k = a + kd \).
\( A_1 = 49 + 1(2) = 51 \)
\( A_{25} = 49 + 25(2) = 49 + 50 = 99 \)
\( A_{47} = 49 + 47(2) = 49 + 94 = 143 \)
\( A_{49} = 49 + 49(2) = 49 + 98 = 147 \)
Mean \( = \frac{A_1 + A_{25} + A_{47} + A_{49}}{4} \)
Mean \( = \frac{51 + 99 + 143 + 147}{4} = \frac{440}{4} = 110 \)
Step 4: Final Answer:
The mean of the specified arithmetic means is 110.